The isomerization reaction H3CNC(g) <=> H3CCN(g) was found to be first order in H3CNC. At 600 K, the concentration of H3CNC is 75% of its original value after 205 s.
(a) What is the rate constant for this isomerization reaction at 600 K?
(b) At what time will 33% of the H3CNC originally present have isomerized?
(c) What is the half-life for this reaction at 600 K?
You need to read about chemical kinetics in your text. Then post what you think your approach should be.
To find the rate constant for the isomerization reaction at 600 K, we can use the first-order rate equation:
ln([A]t/[A]0) = -kt
Where [A]t is the concentration of reactant A at a certain time t, [A]0 is the initial concentration of reactant A, k is the rate constant, and ln is the natural logarithm.
(a) We are given that at 600 K, the concentration of H3CNC is 75% of its original value after 205 s. This means [A]t/[A]0 = 0.75 and t = 205 s. Plugging these values into the equation above, we get:
ln(0.75) = -k * 205
Now we just need to solve for k. Rearranging the equation:
k = -ln(0.75) / 205
Calculating this value will give us the rate constant for the isomerization reaction at 600 K.
(b) To find the time at which 33% of the H3CNC originally present has isomerized, we need to use the same equation and solve for t. This time we have [A]t/[A]0 = 0.33 and we want to find t. The equation becomes:
ln(0.33) = -k * t
Now we rearrange the equation to solve for t:
t = -ln(0.33) / k
Calculating this value will give us the time at which 33% of the H3CNC has isomerized.
(c) The half-life for a first-order reaction is given by the equation:
t1/2 = ln(2) / k
Substituting the value of k obtained in part (a) into this equation will give us the half-life for this reaction at 600 K.